Testing for Uniform Continutity with Weierstrass MTest

[jbunniii]

Evaluate
$$f_n(x) = \frac{x}{1+nx^2}$$
at $x = \frac{1}{\sqrt{n}}$:
$$f_n\left(\frac{1}{\sqrt{n}}\right) = \frac{1/\sqrt{n}}{1 + n(1/\sqrt{n})^2} = ?$$

 

[Emspak]

Oh, SNAP! THat was stupid of me. So, you get $$\frac{1}{2\sqrt{n}}$$

 

[jbunniii]

Right. So now you have shown the following:
$$|f_n(x)| \leq \max_x |f_n(x)| = \frac{1}{2\sqrt{n}}$$
And $\frac{1}{2\sqrt{n}} \rightarrow 0$ as $n \rightarrow \infty$. What can you conclude?

 

[Emspak]

Well, that would point to uniform convergence then wouldn't it? We have a sequence of numbers that approaches zero as n--> infinity. It's also < infinity. And on top of that the absolute value of the function is less than that maximum for all n.

 

[jbunniii]

Well, that would point to uniform convergence then wouldn't it? We have a sequence of numbers that approaches zero as n--> infinity. It's also < infinity. And on top of that the absolute value of the function is less than that maximum for all n.

Yes, that's right. The conclusion is that $f_n$ converges uniformly to $0$.

 

[Emspak]

Thanks for bearing with me. This has cleared a bunch of things up, and it can get a bit frustrating.

 

[jbunniii]

No problem, this stuff is always confusing at first. Just keep doing exercises like this and it will become a lot clearer.